Python tf.linalg.LinearOperatorIdentity用法及代碼示例

LinearOperator 的作用類似於 [batch] 方單位矩陣。

繼承自：LinearOperator，Module

用法

tf.linalg.LinearOperatorIdentity(
    num_rows, batch_shape=None, dtype=None, is_non_singular=True,
    is_self_adjoint=True, is_positive_definite=True, is_square=True,
    assert_proper_shapes=False, name='LinearOperatorIdentity'
)

該運算符的作用類似於 [batch] 單位矩陣 A ，對於某些 b >= 0 ，其形狀為 [B1,...,Bb, N, N] 。第一個 b 索引索引批處理成員。對於每個批次索引 (i1,...,ib) , A[i1,...,ib,::] 是一個 N x N 矩陣。此矩陣A 未具體化，但為了廣播此形狀將是相關的。

LinearOperatorIdentity 使用 num_rows 以及可選的 batch_shape 和 dtype 參數進行初始化。如果 batch_shape 是 None ，則此運算符有效地傳遞所有參數。如果提供了batch_shape，則可能會發生廣播，這將需要製作副本。

# Create a 2 x 2 identity matrix.
operator = LinearOperatorIdentity(num_rows=2, dtype=tf.float32)

operator.to_dense()
==> [[1., 0.]
     [0., 1.]]

operator.shape
==> [2, 2]

operator.log_abs_determinant()
==> 0.

x = ... Shape [2, 4] Tensor
operator.matmul(x)
==> Shape [2, 4] Tensor, same as x.

y = tf.random.normal(shape=[3, 2, 4])
# Note that y.shape is compatible with operator.shape because operator.shape
# is broadcast to [3, 2, 2].
# This broadcast does NOT require copying data, since we can infer that y
# will be passed through without changing shape.  We are always able to infer
# this if the operator has no batch_shape.
x = operator.solve(y)
==> Shape [3, 2, 4] Tensor, same as y.

# Create a 2-batch of 2x2 identity matrices
operator = LinearOperatorIdentity(num_rows=2, batch_shape=[2])
operator.to_dense()
==> [[[1., 0.]
      [0., 1.]],
     [[1., 0.]
      [0., 1.]]]

# Here, even though the operator has a batch shape, the input is the same as
# the output, so x can be passed through without a copy.  The operator is able
# to detect that no broadcast is necessary because both x and the operator
# have statically defined shape.
x = ... Shape [2, 2, 3]
operator.matmul(x)
==> Shape [2, 2, 3] Tensor, same as x

# Here the operator and x have different batch_shape, and are broadcast.
# This requires a copy, since the output is different size than the input.
x = ... Shape [1, 2, 3]
operator.matmul(x)
==> Shape [2, 2, 3] Tensor, equal to [x, x]

形狀兼容性

該運算符作用於具有兼容形狀的 [batch] 矩陣。 x 是與 matmul 和 solve 的形狀兼容的批處理矩陣，如果

operator.shape = [B1,...,Bb] + [N, N],  with b >= 0
x.shape =   [C1,...,Cc] + [N, R],
and [C1,...,Cc] broadcasts with [B1,...,Bb] to [D1,...,Dd]

性能

如果 batch_shape 初始化參數是 None ：

• operator.matmul(x) 是 O(1)
• operator.solve(x) 是 O(1)
• operator.determinant() 是 O(1)

如果提供了batch_shape 初始化arg，並且靜態檢查不能排除廣播的需要：

• operator.matmul(x) 是 O(D1*...*Dd*N*R)
• operator.solve(x) 是 O(D1*...*Dd*N*R)
• operator.determinant() 是 O(B1*...*Bb)

矩陣屬性提示

此 LinearOperator 使用 is_X 形式的布爾標誌初始化，用於 X = non_singular, self_adjoint, positive_definite, square 。它們具有以下含義：

• 如果 is_X == True ，調用者應該期望操作符具有屬性 X 。這是一個應該實現的承諾，但不是運行時斷言。例如，有限的浮點精度可能會導致違反這些承諾。
• 如果 is_X == False ，調用者應該期望操作符沒有 X 。
• 如果is_X == None(默認)，調用者應該沒有任何期望。